Find the roots of z^3=1 using De Moivre theorem
De Moivre's theorem states that for any complex number z = r(cos θ + i sin θ), and any positive integer n,
z^n = r^n (cos nθ + i sin nθ)
In this case, we have z^3 = 1, which means we can write:
z = cos(2πk/3) + i sin(2πk/3), where k = 0, 1, 2
Therefore, the roots of z^3 = 1 are:
z1 = cos(0) + i sin(0) = 1
z2 = cos(2π/3) + i sin(2π/3) = -1/2 + i(√3/2)
z3 = cos(4π/3) + i sin(4π/3) = -1/2 - i(√3/2)
To find the roots of the equation z^3 = 1 using De Moivre's theorem, we can represent 1 as r(cosθ + isinθ), where r = 1 and θ = 0.
Now, let's find the roots step-by-step:
Step 1: Applying De Moivre's theorem
Let z = r^(1/n) * [cos(θ/n) + isin(θ/n)] be the roots of z^n = r * (cosθ + isinθ).
In this case, n = 3, r = 1, θ = 0.
Step 2: Finding the principal root
The principal root (or primary root) is given by:
z₁ = 1^(1/3) * [cos(0/3) + isin(0/3)].
Simplifying this gives:
z₁ = 1 * [cos(0) + i * sin(0)].
Since cos(0) = 1 and sin(0) = 0, z₁ becomes:
z₁ = 1 * (1 + i * 0) = 1.
So, the first root is z₁ = 1.
Step 3: Finding the other roots
We need to find the other two roots. We can do this by adding multiples of 2π to the principal root.
Let's find the second root:
z₂ = 1^(1/3) * [cos((0 + 2π)/3) + i * sin((0 + 2π)/3)].
Simplifying this gives:
z₂ = 1 * [cos(2π/3) + i * sin(2π/3)].
Using the trigonometric identity cos(2π/3) = -1/2 and sin(2π/3) = √3/2, we get:
z₂ = 1 * (-1/2 + i * √3/2) = -1/2 + i * √3/2.
So, the second root is z₂ = -1/2 + i * √3/2.
To find the third root, add another multiple of 2π:
z₃ = 1^(1/3) * [cos((0 + 4π)/3) + i * sin((0 + 4π)/3)].
Simplifying this gives:
z₃ = 1 * [cos(4π/3) + i * sin(4π/3)].
Using the trigonometric identity cos(4π/3) = -1/2 and sin(4π/3) = -√3/2, we get:
z₃ = 1 * (-1/2 - i * √3/2) = -1/2 - i * √3/2.
So, the third root is z₃ = -1/2 - i * √3/2.
Therefore, the roots of the equation z³ = 1 are: 1, -1/2 + i√3/2, and -1/2 - i√3/2.